There exists an element $e \in G$ such that $a * e = a$ and $e * a = a$ (The existence of an identity for $*$).

For all $a \in G$ there exists a $a^{-1} \in G$ such that $a * a^{-1} = e$ and $a^{-1} * a = e$ (The existence of inverses for each element in $G$).

We will now look at the group of continuous real-valued functions on a fixed interval $[a, b]$.

Consider the set of continuous functions $f$ on the interval $[a, b]$ where $a \leq b$, which we denote by $C[a, b]$. Let $+$ be the operation of function addition where for all $f, g \in [0, 1]$ we have that:

(1)

\begin{align} \quad (f + g)(x) = f(x) + g(x) \end{align}

To see if $(C[a, b], +)$ is a group, we must verify that all three points in the definition. From calculus we know that if $f$ and $g$ are continuous on $[a, b]$ then the sum $(f + g)$ is continuous on $[a, b]$ so $C[a, b]$ is closed under function addition.

Therefore $e(x) = 0$ is our identity element. Lastly, recall from calculus that if $f$ is continuous on $[a, b]$ then for any $k \in \mathbb{R}$ we have that $kf$ is continuous on $[a, b]$. If $k = -1$ then $-f$ is continuous on $[a, b]$. So for all $f \in C[a, b]$ we have that $-f \in C[a, b]$ and: